Basic Ajax with jQuery - Execute PHP File

Question

For some reason I can't get this to work. All I want to do is that when a button is clicked, (via ajax) a php file is executed. Aka, if I weren't using ajax, we'd be looking at:

<a href="file.php">dfgdfg</a>

I want the file to be executed without leaving the page.

This is what I've got atm:

$(".vote-up").live('click', function(e) {
$.ajax("file.php", function(){
       alert("Executed file");
   });
});

That doesn't seem to be working. I'm just really confused how the jQuery ajax function functions in general, without dealing with anything remotely complicated.

Added question:

.ajax({
       url: 'includes/login-check-jquery.php',
       success: function (data) {
            if(data == 1){
                alert("Logged in!!!");
            }else{
                window.location = data;
            }
        error: function (xhr, status, error) {
        // executed if something went wrong during call
        if (xhr.status > 0) alert('Error: ' + status); // status 0 - when load is interrupted
        }
    });
});

In the above code, if the data returned equals "logged in" than a message box appears. Otherwise it'll redirect to the location (that is sent in data). For some reason this if statement isn't working, but the data is being returned as expected.

My PHP code is:

<?php  
if (!$user->logged_in){
    echo "../login/index.php";
}else{
    echo "1";
}

?>

Any ideas?

try replacing live with on because live is deprecated ! also check that the path to the php file is correct — unloco
– unloco, Commented Aug 9, 2012 at 2:33
This doesn't look wrong to me--are you trying to get a response from the PHP file or just set something off and forget about it? — Ian Hunter
– Ian Hunter, Commented Aug 9, 2012 at 2:35
I don't believe your anonymous function will be executed without getting a non-error status code from your web server. You can use something like Firebug or Chrome's built-in developer tools' network tab to see the status code from the request. — Ian Hunter
– Ian Hunter, Commented Aug 9, 2012 at 2:37
Look into window.location: developer.mozilla.org/en-US/docs/DOM/window.location — Ian Hunter
– Ian Hunter, Commented Aug 9, 2012 at 2:55
Check @therao's answer--and you might want to ensure you're either passing something back in plaintext or as JSON. — Ian Hunter
– Ian Hunter, Commented Aug 9, 2012 at 3:00

i-- · Accepted Answer · 2012-08-09 03:06:09Z

6

If you don't want to leave the page, do

$(".vote-up").live('click', function(e) {
    e.preventDefault();
    ...

And ajax could be updated, if desired to

$.ajax({
    url: 'file.php',
    success: function (data) {
        // this is executed when ajax call finished well
        alert('content of the executed page: ' + data);
    },
    error: function (xhr, status, error) {
        // executed if something went wrong during call
        if (xhr.status > 0) alert('got error: ' + status); // status 0 - when load is interrupted
    }
});

error callback part could be removed, if you go for simplicity and not usability and some more options could be added by referencing jquery ajax doc.

edited Aug 9, 2012 at 3:06

answered Aug 9, 2012 at 2:59

i--

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10 Comments

Andrew Over a year ago

I'm a bit confused with this. Is data a variable that is assigned the value of whatever is returned by the php file?

Andrew Over a year ago

From testing it, yes that is the case. Thanks :) This'll work :)

i-- Over a year ago

yes, it is. sorry for quick replies without really finishing reading ;)

Andrew Over a year ago

All good. I understand it and I'm getting it working so all good :)

Andrew Over a year ago

Put up a question at the end if you have any ideas :)

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